[Aetos]

As I've mentioned in my last 2 posts I have NOT studied chemistry at A level and so am struggling  particularily with this topic because it is completely new to me. Please try to put it in simple terms if possible.

The free energy change associated with the complete oxidation of glucose

C6H12O6 (s) +  6O2 (g)      →      6CO2 (g)   +   6H2O (l)

Is -3089.0 kJ mol-1.  What is the equilibrium constant for this reaction at 37oC?

[Dan]

http://en.wikipedia.org/wiki/Gibbs_free_energy#Useful_identities

[Aetos]

that does not help at all. Do you not think that I would have attempted to research into how to working it out myself?
The reason I posted in a forum was so that I could get help in understanding how to work it out and not just been thrown a link.
Plus there are so many equations in that link, how do i know which one to apply in this case? And I dont see a particular equation that says it works out equilibrium constant.
Please understand that I am only a beginner and would appreciate proper help where someone could spare a few minutes of their time to explain the above question to me properly and NOT just throw me a link 

[Dan]

To be honest, from the question I did assume you hadn't researched it at all - you provided no indication of an attempt at the question. We get an enormous amount of threads in which the poster has made no attempt to answer the question, so I would say it was a reasonable assumption. It was nothing personal.

there are so many equations in that link, how do i know which one to apply in this case?

You want to find a relationship between :delta: G and K, the Gibbs free energy and the equilibrium constant. The article states:

Moreover, we have,

K = e^{-(:delta: G/RT)}
...
which relates the equilibrium constant with Gibbs free energy.

Seeing as the information you needed was already available online in a digestible form, I did not see a need to re-type it for you.

Furthermore, it links to the Wikipedia article about chemical equilibrium, which explains the topic further, and includes the derivation from first principles. Sorry to provide another link, I'm not forcing you to click it, but I think you may find it useful (depending on how much calculus you've done):

http://en.wikipedia.org/wiki/Chemical_equilibrium

Key equations, such as K = e^{-(:delta: G/RT)} more often  than not just have to be memorised. You can of course work it out from scratch, but I was assuming from the level of the original question that the derivation would be too advanced.

[Aetos]

I provided no indication of attempt because this is the FIRST time I've come across a question like this. I did not study chemistry in college and again will repeate this is my FIRST time. If you just provide me a link it doesnt help because i dont know what parts i need to apply to this particular question. But this doesnt mean that didnt research it or try to understand it myself. But because I found difficulty doing so I decided to seek help in this forum.

I did read through some of it but failed to make sense of it.

it would have been much simpler to say you need to apply this equation:
 K = e-(delta G/RT)

and explain what i need to subsitute into this equation, where and why. but then again perhaps it isnt that simple. Im not looking to understand it fully because i doubt i ever will without proper tutoring, since i havnt done this before.

okay so going back to the equation

I understand that I am trying to work out K?

what does the e- stand for? what do i Subsitute into this part of the equation and why?

Im guessing that -3089.0 kJ mol-1 is subsituted into 'delta G' part of the equation, please correct me if im wrong. also is any conversion needed?

I understand R is a constant but how do you work out R? or what is the value for R? is it always the same?

T is for temperature im guessing which would be 37*C converted into kelvin? is this right?

thanks

[Dan]

I understand that I am trying to work out K?

Correct.

what does the e- stand for? what do i Subsitute into this part of the equation and why?

e is a mathematical constant. It is the inverse of a natural logarithm (ln). It is used in exponential functions with the general equation e^x - there is a button for it on your calculator. As the inverse of a natural logarithm, e^{ln x} = x.

In the context of this question, the equation comes from the equation:
:delta: G = -RT ln K

which can be rearranged:

-(:delta: G/RT) = ln K
e^{-(:delta: G/RT)} = e^{ln K}
e^{-(:delta: G/RT)} = K

More information here: http://en.wikipedia.org/wiki/Exponential_function
and here: http://en.wikipedia.org/wiki/Natural_logarithm

If the concept of logarithms and exponential functions are totally new to you, I would strongly recommend a course in mathematics before continuing with chemistry.

Im guessing that -3089.0 kJ mol-1 is subsituted into 'delta G' part of the equation

Yes, :delta: G is the symbol for "change in Gibbs free energy". You should convert to SI units, which for energy change is Joules per mole.

I understand R is a constant but how do you work out R? or what is the value for R? is it always the same?

R is the universal gas constant. It's actual value depends on which units you are working in - but this is almost always SI units where R = 8.314 JK^-1mol^-1

T is for temperature im guessing which would be 37*C converted into kelvin? is this right?

Yes, if you use SI units for the gas constant, then you need to convert to Kelvin (the SI unit of temperature). The units must match for each component:

the term:

-(:delta: G/RT)

should be unitless - all the units should cancel.

if we measure energy in J mol^-1, gas constant in J K^-1 mol^-1 and temperature in K, considering the units only we get:

J mol^-1/(J K^-1 mol^-1 K)

I have colour coded the terms that cancel, overall everything cancels and we have no units.

[Aetos]

Okay trying to take it one step at a time. so would it be worked out like this

K = e-(delta G/RT)

convert -3089.0 kJ mol-1 to J mol-1 = -3089 x 1000= -3089000

convert R constant 8.314 JK-1mol-1 to J-1mol-1 = 8314

convert 37*C to kelvin= 310

then subsitute values into equation

K= e-(3089000/(8314x310))


Is that right so far?


and I still dont know how to work out e- can someone show me how to work it out like I've done above. I feel like giving up because i still dont understand it.

Note: im in university studying forensics, I havent studied chemistry or maths at A level so it would be helpful if you just showed me the working out with the answer as you are explaining it, that way i have a better chance of understanding it. And incase your afraid to tell me the answer just note that this question was derived from a worksheet that was completely optional. I am going through the aggravation of trying to understand even a tiny bit of it for my own benefit. so please help a dumb girl out. thanks

[Aetos]

okay  you said once u rearrange the equation i'd get

e- ^ (delta G/RT) = K

and that way i can work out the constant

and e- is actually inverse log so in a calculator i would put in:

10 to the power (-3089000/(8314x310)) =0.0633

so K = 0.0633

Is that right? and did i do the conversion right because i feel like the numbers look wrong >.<

[UG]

10 to the power (-3089000/(8314x310)) =0.0633

so K = 0.0633

Is that right? and did i do the conversion right because i feel like the numbers look wrong >.<

The gas constant value R is actually 8.314 JK^-1 mol^-1 not 8314
Your final rearranged equation (Using the method posted by Dan) should be K = e^{-(-3089000/(8.314x310))} which gives K = e^1198.5 and it is such a large number that my calculator runs out of space 

[Borek]

and e- is actually inverse log so in a calculator i would put in:

10 to the power

These are different logarithms. One is base 10 log, the other is base e log. 10^x and e^x are different numbers.

http://en.wikipedia.org/wiki/Logarithm

[Dan]

okay  you said once u rearrange the equation i'd get

e- ^ (delta G/RT) = K

No, the minus is in the exponent, e^ -(:delta: G/RT),

and e- is actually inverse log so in a calculator i would put in:

10 to the power (-3089000/(8314x310)) =0.0633

e is an inverse natural logarithm, inverse ln, not inverse log.

It is e to the power -(-3089000/(8.314x310))

There is an e^x button on your calculator

so K = 0.0633

Is that right? and did i do the conversion right because i feel like the numbers look wrong >.<

No that number is not sensible. Consider what an equilibrium constant actually is, K = [products]/[reactants]

Since the combustion of ethane is practically irreversible, we would expect this value to be very very large.

im in university studying forensics, I havent studied chemistry or maths at A level

If I were you, I'd get some A-level chemistry and maths textbooks and teach yourself, or even better get someone to give you some tuition. Postgrads at your University will probably give you one on one tuition at a good rate.

Unless your university runs catch-up chemistry and maths courses, I think it was outrageously irresponsible of them to take you on for forensics. You are going to struggle very seriously if you don't have a grounding in chemistry and maths. I'm sure you can manage, but it's going to be tough - the sooner you can catch up some A-level maths the better.

I would get:
http://www.amazon.co.uk/Edexcel-AS-Level-Modular-Mathematics/dp/0435519107
and
http://www.amazon.co.uk/Edexcel-AS-Level-Modular-Mathematics/dp/0435519115

[Aetos]

thanks everyone for your help 

I think i understand how to work it out more or less.

[Aetos]

If I were you, I'd get some A-level chemistry and maths textbooks and teach yourself, or even better get someone to give you some tuition. Postgrads at your University will probably give you one on one tuition at a good rate.

Unless your university runs catch-up chemistry and maths courses, I think it was outrageously irresponsible of them to take you on for forensics. You are going to struggle very seriously if you don't have a grounding in chemistry and maths. I'm sure you can manage, but it's going to be tough - the sooner you can catch up some A-level maths the better.

I would get:
http://www.amazon.co.uk/Edexcel-AS-Level-Modular-Mathematics/dp/0435519107
and
http://www.amazon.co.uk/Edexcel-AS-Level-Modular-Mathematics/dp/0435519115

I'm actually thankful that I got into doing  forensics at university without those maths and science Alevels, it's pretty brain rattling but I can see how it would've helped. However I studied Forensic science at A level so I've been okay with everything else whether it be biology or the other chemistry topics so far. I'm just hoping I get through the first year with flying colours although it doesnt really apply to my final degree. So wish me luck! and thanks again!